Acid - Base
1970
(a) What is the pH of a 2.0 molar solution of acetic
acid. Ka acetic acid = 1.8×10–5
(b) A buffer solution is prepared by adding 0.10
liter of 2.0 molar acetic acid solution to 0.1 liter
of a 1.0 molar sodium hydroxide solution.
Compute the hydrogen ion concentration of the
buffer solution.
(c) Suppose that 0.10 liter of 0.50 molar hydrochloric acid is added to 0.040 liter of the buffer prepared in (b). Compute the hydrogen ion concentration of the resulting solution.
Answer:
(a) CH3COOH(aq) √ H+(aq) + CH3COO–(aq)
Ka = 1.8 ×10
−5
[H+ ][CH3COO− ]
=
[CH3COOH]
page 1
1970
H3PO2,H3PO3, and H3PO4 are monoprotic, diprotic
and triprotic acids, respectively, and they are about
equal strong acids.
HClO2, HClO3, and HClO4 are all monoprotic acids,
but HClO2 is a weaker acid than HClO3 which is
weaker than HClO4. Account for:
(a) The fact that the molecules of the three
phosphorus acids can provide different numbers
of protons.
(b) The fact that the three chlorine acids differ in
strengths.
Answer:
(a) the structure for the three acids are as follows:
O
H
H
H P O O P O H H O P O H
O
O
O
[H+] = [CH3COO–] = X
H
H
H
[CH3COOH] = 2.0 – X, X << 2.0, (2.0– X) = 2.0
The hydrogen atom(s) bonded directly to the
X2
–5
phosphorus atom is/are not acidic in aqueous so1.8×10 =
2.0
lution; only those hydrogen atoms bonded to the
X = 6.0x10–3 = [H+]; pH = –log [H+] = 2.22
oxygen atoms can be released as protons.
(b) 0.1 L x 2.0 mol/L = 0.20 mol CH3COOH
(b) The acid strength is successively greater as the
number of oxygen atoms increases because the
0.1 L x 1.0 mol/L = 0.10 mol NaOH
very electronegative oxygen atoms are able to
the 0.10 mol of hydroxide neutralizes 0.10 mol
draw electrons away from the chlorine atom and
CH3COOH with 0.10 mol remaining with a conthe O–H bond. This effect is more important as
centration of 0.10 mol/0.20 L = 0.5 M. This also
the number of attached oxygen atoms increases.
produces 0.10 mol of acetate ion in 0.20 L,
This means that a proton is most readily
therefore, [CH3COO–] = 0.50 M.
produced by the molecule with the largest
+ ][0.50]
[H
number of attached oxygen atoms.
−5
1.8 ×10 =
[0.50]
1970
[H+] = 1.8x10–5 = pH of 4.74
A comparison of the theories Arrhenius, Brönsted
(c) [CH3COOH]o = [CH3COO–]o
and Lewis shows a progressive generalization of the
0.040 L
acid base concept. Outline the essential ideas in each
= 0.50 M x
= 0.143 M
of these theories and select three reactions, one that
0.14 L
can be interpreted by all three theories, one that can
0.10 L
[H+]o = 0.50 M x
= 0.357 M
be interpreted by two of them, and one that can be
0.14 L
the equilibrium will be pushed nearly totally to interpreted by only one of the theories. Provide these
the left resulting in a decrease of the hydrogen six interpretations.
ion by 0.143M. Therefore, the [H+]eq = 0.357M Answer:
Arrhenius
– 0.143M = 0.214M.
acid = produce H+ ions in aqueous solution
base = produce OH– ions in aqueous solution
Brönsted–Lowry
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Acid - Base
(H+)
acid = proton
donor; base = proton acceptor
Lewis
acid = e- pair acceptor; base = e- pair donor
page 2
Examples:
Interpreted by all three
HCl + H2O → H+(aq) + Cl–(aq)
NaOH + H2O → Na+(aq) + OH–(aq)
Interpreted by two
NH3 + HCl √ NH4+ + Cl–
Interpreted by only one
BF3 + NH3 → F3B:NH3
1972 (repeated in gases topic)
A 5.00 gram sample of a dry mixture of potassium
hydroxide, potassium carbonate, and potassium chloride is reacted with 0.100 liter of 2.00 molar HCl solution
(a) A 249 milliliter sample of dry CO2 gas, measured at 22˚C and 740 torr, is obtained from this
reaction. What is the percentage of potassium
carbonate in the mixture?
(b) The excess HCl is found by titration to be
chemically equivalent to 86.6 milliliters of 1.50
molar NaOH. Calculate the percentages of
potassium hydroxide and of potassium chloride
in the original mixture.
Answer:
(a) K2CO3 + 2 HCl → CO2 + 2 KCl + H2O
PV
(740torr)(249mL)
mol CO2 =
=
=
RT 62400 m L ⋅torr (295K)
(
mol⋅K
)
= 0.0100 mol CO2
0.10mol CO2 1mol K2 CO3 138.2g K2 CO3
×
×
1mol CO2
1mol K2 CO3
= 1.38 g
1.38gK2 CO3
× 100% = 27.6% K2CO3
5.00 g mixture
(b) orig. mol HCl = 0.100 L × 2.00M = 0.200 mol
reacted with K2CO3
= 0.020 mol
excess HCl = 0.0866L × 1.50M
= 0.130 mol
mol HCl that reacted w/KOH
= 0.050 mol
0.050 mol KOH = 2.81 g = 56.1% of sample
the remaining KCl amounts to 16.3%
1972
Given a solution of ammonium chloride. What additional reagent or reagents are needed to prepare a
buffer from the ammonium chloride solution?
Explain how this buffer solution resists a change in
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Acid - Base
pH when:
(a) Moderate amounts of strong acid are added.
(b) Moderate amounts of strong base are added.
(c) A portion of the buffer solution is diluted with
an equal volume of water.
page 3
Answer:
Since ammonium chloride is a salt of a weak base,
the weak base is needed, ammonia, NH3.
(a) When moderate amounts of a strong acid, H+,
are added, the ammonia reacts with it. The concentration of the hydrogen ion remains
essentially the same and therefore only a very
small change in pH.
NH3 + H+ √ NH4+
(b) When moderate amounts of a strong base, OH–,
are added, the ammonium ion reacts with it. The
concentration of the hydrogen ion remains
essentially the same and therefore only a very
small change in pH.
NH4+ + OH– √ NH3 + H2O
(c) By diluting with water the relative concentration
ratio of [NH4+]/[NH3] does not change, therefore there should be no change in pH.
1973
A sample of 40.0 milliliters of a 0.100 molar
HC2H3O2 solution is titrated with a 0.150 molar
NaOH solution. Ka for acetic acid = 1.8×10–5
(a) What volume of NaOH is used in the titration in
order to reach the equivalence point?
(b) What is the molar concentration of C2H3O2– at
the equivalence point?
(c) What is the pH of the solution at the equivalence
point?
Answer:
(a) MaVa=MbVb
(0.100M)(40.0 mL) = (0.150M)(Vb)
Vb = 26.7 mL
(b) acetate ion is a weak base with
K
1.0 × 10 −14
Kb = w =
= 5.6 × 10−10
−5
Ka
1.8 × 10
4.00mmol
[CH3COO− ]o =
= 0.0600M
(40.0mL + 26.7mL)
[CH3COO–]eq = 0.600M –X
[OH–] = [CH3COOH] = X
X2
; X = 9.66 × 10 −5 M
(0.0600 − X)
0.0600M–9.66×10–5M = 0.0599M [CH3COO–]eq
5.6 × 10−10 =
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Acid - Base
1.0 × 10−14
Kw
= 1.04 × 10 −10 M
− =
−5
[OH ] 9.66 × 10
pH = –log [H+] = –log(1.04×10–10) = 9.98
1974 A
A solution is prepared from 0.0250 mole of HCl,
0.10 mole propionic acid, C2H5COOH, and enough
water to make 0.365 liter of solution. Determine the
concentrations of H3O+, C2H5COOH, C2H5COO–,
and OH– in this solution. Ka for propionic acid =
1.3×10–5
Answer:
C2H5COOH + H2O √ C2H5COO– + H3O+
+
(c) [H ] =
page 4
(0.50 − X)(0.100 − X)
; X = 0.100M
(0.50 + X)
using the Henderson–Hasselbalch equation
[NH+ ]
 0.40
4 
 = 4.57
pH = pKa + log
= 4.74 + log
 0.60
 [NH3]
1.8 × 10−5 =
[C2H5COO–] = X
[C2H5COOH] = (0.10mol/0.365L) – X
[H3O+] = (0.0250mol/0.365L) + X
(X)(0.0685 + X)
−5
−5
Ka =
=1.3 ×10 ; X = 5.2 ×10
0.274 − X
[C2H5COO–] = 5.2×10–5M; [C2H5COOH] = 0.274M
[H3O+] = 0.0685M;
Kw
[OH− ] =
= 1. 46 × 10 −13 M
0.0685
1975 A
(a) A 4.00 gram sample of NaOH(s) is dissolved in
enough water to make 0.50 liter of solution. Calculate the pH of the solution.
(b) Suppose that 4.00 grams of NaOH(s) is dissolved
in 1.00 liter of a solution that is 0.50 molar in
NH3 and 0.50 molar in NH4+. Assuming that
there is no change in volume and no loss of NH3
to the atmosphere, calculate the concentration of
hydroxide ion, after a chemical reaction has occurred. [Ionization constant at 25˚C for the reaction NH3 + H2O → NH4+ + OH–; K = 1.8×10–5]
Answer:
4.00gNaOH 1mol
×
= 0.20M
(a)
0.50L
40.0g
K
[H+ ] = w = 5 × 10−14 ; pH = − log[H+ ] = 13.3
0.20
0.100mol
−
(b) [OH ] =
−X
1.00L
[NH4+] = 0.50M – X; [NH3] = 0.50M + X
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Acid - Base
page 5
1975
Reactions requiring either an extremely strong acid
or an extremely strong base are carried out in
solvents other than water. Explain why this is
necessary for both cases.
Answer:
Water is amphoteric and can behave either as an acid
in the presence of a strong base or as a base in the
presence of strong acid.
Water also undergoes autoionization.
1977
The value of the ionization constant, Ka, for
hypochlorous acid, HOCl, is 3.1×10–8.
(a) Calculate the hydronium ion concentration of a
0.050 molar solution of HOCl.
(b) Calculate the concentration of hydronium ion in
a solution prepared by mixing equal volumes of
0.050 molar HOCl and 0.020 molar sodium
hypochlorite, NaOCl.
(c) A solution is prepared by the disproportionation
reaction below. Cl2 + H2O → HCl + HOCl
1976
Calculate the pH of the solution if enough chloH2S + H2O √ H3O+ + HS–
K1 = 1.0×10–7
rine is added to water to make the concentration
HS– + H2O √ H3O+ + S2–
K2 = 1.3×10–13
of HOCl equal to 0.0040 molar.
+
2–
–20
H2 S + 2 H 2 O √ 2 H3 O + S
K = 1.3×10
Answer:
+
2–
–51
Ag2S(s) √ 2 Ag + S
Ksp= 5.5×10
(a) HOCl + H2O √ H3O+ + OCl–
+
−
(a) Calculate the concentration of H3O+ of a soluX2
−8 [H3O ][OCl ]
3.2 × 10 =
=
tion which is 0.10 molar in H2S.
[HOCl]
(0.050 − X)
(b) Calculate the concentration of the sulfide ion,
X = [H3O+] = 4.0×10–5M
S2–, in a solution that is 0.10 molar in H2S and
(b) HOCl + H2O √ H3O+ + OCl–
+
0.40 molar in H3O .
[H3O+ ][0.010 + X]
= 3.2 × 10 −8 ; X ‹‹ 0.010
(c) Calculate the maximum concentration of silver
[0.0250 − X]
ion, Ag+, that can exist in a solution that is
X = [H3O+] = 8.0×10–8M
1.5×10–17 molar in sulfide ion, S2–.
(c) Cl2 + H2O → HCl + HOCl
Answer:
(a) H2S + H2O √ H3O+ + HS–
[HOCl] = [HCl] = 0.0040M
HCl as principal source of H3O+
[H3O+ ][HS− ]
−7
= 1.0 × 10
pH = –log[H3O+] = 2.40
[H2 S]
let X = [H3O+] = [HS–]
1978 A
A 0.682 gram sample of an unknown weak monoprotic organic acid, HA was dissolved in sufficient
water to make 50 milliliters of solution and was
titrated with a 0.135 molar NaOH solution. After the
(b) H2S + 2 H2O √ 2 H3O+ + S2–
addition of 10.6 milliliters of base, a pH of 5.65 was
+
2
2−
recorded. The equivalence point (end point) was
[H3O ] [S ]
= 1.3 × 10 −20
reached after the addition of 27.4 milliliters of the
[H2S]
0.135 molar NaOH.
[0.40]2[S2− ]
−20
= 1.3 × 10 ; [S2–] = 8.1×10–21M (a) Calculate the number of moles of acid in the
[0.10]
original sample.
(c) Ag2S(s) √ 2 Ag+ + S2–; [Ag+]2[S2–] = 5.5×10–51 (b) Calculate the molecular weight of the acid HA.
(c) Calculate the number of moles of unreacted HA
5.5 × 10 −51
−17
remaining in solution when the pH was 5.65.
[Ag + ] =
=
1.9
×
10
M
1.5 × 10 −17
(d) Calculate the [H3O+] at pH = 5.65
X2
= 1.0 × 10−7
0.10 − X
X ‹‹ 0.10; X = 1.0×10–4 M = [H3O+]
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Acid - Base
page 6
(e) Calculate the value of the ionization constant, Answer:
Ka, of the acid HA.
(a) at equivalence point, moles HA = moles NaOH
= MbVb = (0.0274 L)(0.135 M) = 3.70×10–3 mol
HA
mass HA
0.682g
=
= 184 g/mol
(b) molec.wt . =
mol HA
3.70×10−3 mol
(c) HA + OH– → A– + H2O
initial:
0.00370 mol
added:
(0.0106L)(0.135M) = 0.00143 mole
remaining: (0.00370 – 0.00143) = 0.00227 mol
(d) pH = –log[H3O+]; [H3O+] = 10–pH = 10–5.65
= 2.2×10–6M
[H3O + ][A− ] (2.2 × 10 −6 )(0.00143 / v)
=
[HA]
(0.00227 / v)
–6
= 1.4×10
(e) Ka =
1978 D
Predict whether solutions of each of the following
salts are acidic, basic, or neutral. Explain your
prediction in each case
(a) Al(NO3)3
(b) K2CO3
(c) NaBr
Answer:
(a) acidic; Al3+ + H2O √ AlOH2+ + H+;
hydrolysis of Al3+;
Al(OH2)n3+ as Brönsted acid, etc.
(b) basic; CO32– + H2O √ HCO3– + OH– ; or
hydrolysis of CO32– as conjugate to a weak
acid, etc.
(c) neutral; Na+ from strong base; Br– from strong
acid
1979 B
A solution of hydrochloric acid has a density of 1.15
grams per milliliter and is 30.0% by weight HCl.
(a) What is the molarity of this solution of HCl?
(b) What volume of this solution should be taken in
order to prepare 5.0 liters of 0.20 molar hydrochloric acid by dilution with water?
(c) In order to obtain a precise concentration, the
0.20 molar hydrochloric acid is standardized
against pure HgO (molecular weight = 216.59)
by titrating the OH– produced according to the
following quantitative reaction.
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Acid - Base
HgO(s) + 4
I–
+ H2O → HgI4
2–
+2
OH–
page 7
In a typical experiment 0.7147 grams of HgO required 31.67 milliliters of the hydrochloric acid
solution for titration. Based on these data what is
the molarity of the HCl solution expressed to
four significant figures.
Answer:
1.15g 1000mL 30.0gHCl 1mol
×
×
×
= 9.5M
(a)
1mL
1L
100g
35.5g
(b) MfVf = MiVi
(0.20M)(5.0L) = (9.5M)(V)
V = 0.11 L
0.7147g
= 0.003300 mol HgO
(c)
216.59g mol
mol OH– prod. = 2 (mol HgO) = 0.006600 mol
mol HCl req. = mol OH– prod. = 0.006600 mol
0.006600mol
M HCl =
= 0.2084M
0.03167L
1979 D
NH4+ + OH– √ NH3 + H2O
H2O + C2H5O– √ C2H5OH + OH–
The equations for two acid–base reactions are given
above. Each of these reactions proceeds essentially
to completion to the right when carried out in
aqueous solution.
(a) Give the Brönsted–Lowry definition of an acid
and a base.
(b) List each acid and its conjugate base for each of
the reactions above.
(c) Which is the stronger base, ammonia or the
ethoxide ion. C2H5O–? Explain your answer.
Answer:
(a) acid = proton donor; base = proton acceptor
(b)
Acid
Conjugate base
1st reaction
NH4+
NH3
H2 O
OH–
2nd reaction
H2 O
OH–
C2H5OH
C2H5O–
(c) ethoxide is a stronger base than ammonia.
A stronger base is always capable of displacing
a weaker base. Since both reactions are quantitative, in terms of base strength, OH– > NH3 in 1st
reaction; C2H5O– > OH– in 2nd rxn.
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teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
page 8
1980 A
Methylamine CH3NH2, is a weak base that ionizes in
solution as shown by the following equation.
CH3NH2 + H2O √ CH3NH3+ + OH–
(a) At 25˚C the percentage ionization in a 0.160
molar solution of CH3NH2 is 4.7%. Calculate
[OH–], [CH3NH3+], [CH3NH2], [H3O+], and the
pH of a 0.160 molar solution of CH3NH2 at
25˚C
(b) Calculate the value for Kb, the ionization constant for CH3NH2, at 25˚C.
(c) If 0.050 mole of crystalline lanthanum nitrate is
added to 1.00 liter of a solution containing 0.20
mole of CH3NH2 and 0.20 mole of its salt
CH3NH3Cl at 25˚C, and the solution is stirred
until equilibrium is attained, will any La(OH)3
precipitate? Show the calculations that prove
your answer. (The solubility constant for
La(OH)3, Ksp = 1×10–19 at 25˚C)
Answer:
(a) CH3NH2; 0.160M × 4.7% = 7.5×10–3 M ionizing
(0.160M – 0.0075M) = 0.0152M @ equilibrium
[CH3NH3+] = [OH–] = 7.5×10–3 M
Kw
−12
[H3O+ ] =
M
−3 = 1.3 × 10
7.5 × 10
pH = –log [H3O+] = 11.89
(b) K b =
[CH3 NH+3 ][OH− ] (7.5×10−3 )2
=
[CH3 NH2 ]
0.152
=3.7×10–4
−4
(c) K b = 3.7 × 10 =
(0.20 + X)(X)
≈X
(.020 − X)
= [OH–]
Q = [La3+][OH–]3 = (0.050)(3.7×10–4)3
= 2.5×10–12
Q > Ksp, therefore, La(OH)3 precipitates
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1981 D
Al(NO3)3
Acid - Base
page 9
(b) The pH remains unchanged because the ratio of
the formate and formic acid concentration stays
the same.
(a) Predict whether a 0.10 molar solution of each of
the salts above is acidic, neutral or basic.
(b) For each of the solutions that is not neutral,
write a balanced chemical equation for a
reaction occurring with water that supports your
prediction.
Answer:
(a) Al(NO3)3 – acidic
K2CO3 – basic
K2CO3 NaHSO4
NaHSO4 – acidic
NH4Cl
NH4Cl – acidic
(b) Al3+ + H2O → Al(OH)2+ + H+
Al(H2O)63+ + H2O → [Al(H2O)5OH]2+ + H3O+
Al3+ + 3 H2O → Al(OH)3 + 3 H+
CO32– + H2O → HCO3– + OH–
HSO4– + H2O → SO42– + H3O+
NH4+ + H2O → NH3 + H3O+
1982 A
A buffer solution contains 0.40 mole of formic acid,
HCOOH, and 0.60 mole of sodium formate,
HCOONa, in 1.00 litre of solution. The ionization
constant, Ka, of formic acid is 1.8×10–4.
(a) Calculate the pH of this solution.
(b) If 100. millilitres of this buffer solution is
diluted to a volume of 1.00 litre with pure water,
the pH does not change. Discuss why the pH
remains constant on dilution.
(c) A 5.00 millilitre sample of 1.00 molar HCl is
added to 100. millilitres of the original buffer
solution. Calculate the [H3O+] of the resulting
solution.
(d) A 800.–milliliter sample of 2.00–molar formic
acid is mixed with 200. milliliters of 4.80–molar
NaOH. Calculate the [H3O+] of the resulting solution.
Answer:
(a) using the Henderson–Hasselbalch equation
 [A − ] 
pH = pK a +log

[HA]
 0.60 
= − log(1.8×10−4 )+ log

 0.40 
= 3.92
{other approaches possible}
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Acid - Base
page 10
Conductivity, λ
(c) initial concentrations
hydroxide originally present in the solution that
is titrated.
5.00mL
1.00M HCl ×
= 0.0476M
(d) Explain why the conductivity does not fall to
105mL
zero at the equivalence point of this titration.
100mL
0.40M HCOOH×
= 0.38M
Answer:
105mL
100mL
(a) Ba2+ + 2 OH– + 2 H+ + SO42– → BaSO4(s) + 2
0.60M HCOO- ×
= 0.57M
105mL
H2 O
+
–
concentrations after H reacts with HCOO
(b) The initial conductivity is high because of the
0.38M + 0.05M = 0.43M HCOOH
presence of Ba2+ and OH– ions. The
conductivity decreases because Ba2+ forms
0.57M – 0.05M = 0.52M HCOO–
insoluble BaSO4 with the addition of SO42–.
0.43M
[H3O+ ] = 1.8 × 10 −4 ×
= 1.5 × −4 M
The conductivity also decreases because OH–
0.52M
combines with the addition of H+ ions by
(d) 0.800L × 2.00M HCOOH = 1.60 mol
forming H2O.
0.200L × 4.80M NaOH = 0.96 mol OH–
Beyond the equivalence point conductivity inat equil., (1.60 – 0.96) = 0.64 mol HCOOH and
creases as H+ and SO42– ions are added.
0.96 mol HCOO–
(c) # mol Ba(OH)2 = # mol H2SO4
0.64M
[H3O+ ] = 1.8 × 10 −4 ×
= 1.2 × −4 M
0.96M
=0.1M × 0.04L = 0.004 mol
(d) BaSO4(s) dissociates slightly to form Ba2+ and
1982 D
SO42–, while the water ionizes slightly to form
A solution of barium hydroxide is titrated with 0.1–
H+ and OH–.
M sulfuric acid and the electrical conductivity of the
solution is measured as the titration proceeds. The
1983 B
data obtained are plotted on the graph below.
The molecular weight of a monoprotic acid HX was
to be determined. A sample of 15.126 grams of HX
was dissolved in distilled water and the volume
brought to exactly 250.00 millilitres in a volumetric
flask. Several 50.00 millilitre portions of this
solution were titrated against NaOH solution, requiring an average of 38.21 millilitres of NaOH.
The NaOH solution was standardized against oxalic
acid dihydrate, H2C2O4.2H2O (molecular weight:
126.066 gram mol–1). The volume of NaOH solution
required to neutralize 1.2596 grams of oxalic acid dihydrate was 41.24 millilitres.
10 20 30 40 50 60 70 80
(a) Calculate the molarity of the NaOH solution.
Millilitres of 0.1–M H2SO4
(b) Calculate the number of moles of HX in a 50.00
millilitre portion used for titration.
(a) For the reaction that occurs during the titration
described above, write a balanced net ionic (c) Calculate the molecular weight of HX.
equation.
(d) Discuss the effect of the calculated molecular
weight of HX if the sample of oxalic acid dihy(b) Explain why the conductivity decreases, passes
drate contained a nonacidic impurity.
through a minimum, and then increases as the
volume of H2SO4 added to the barium Answer:
hydroxide is increased.
(c) Calculate the number of moles of barium
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Acid - Base
page 11
wrong with the erroneous procedures.
(No calculations are necessary, but the following
126.066
acidity constants may be helpful: acetic acid,
9.9916×10–3 mol
Ka= 1.8×10–5; NH4+, Ka = 5.6×10–10)
H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O
Answer:
2molNaOH
9.9916 × 10 −3 mol ×
= 1.9983 × 10−2 mol(a) A buffer solution resists changes in pH upon the
1molH 2C 2O4
addition of an acid or base.
1.9983 × 10 −2 mol
Preparation of a buffer: (1) mix a weak acid + a
M NaOH =
= 0.4846M
salt of a weak acid; or (2) mix a weak base + salt
0.04124L
of a weak base; or (3) mix a weak acid with
(b) mol HX = mol NaOH
about half as many moles of strong base; or (4)
0.03821 L × 0.4846 M = 0.01852 mol HX
mix a weak base with about half as many moles
0.01852mol
(c)
× 250.00mL = 0.09260mol HX
of strong acid; or (5) mix a weak acid and a
50.00mL
weak base.
15.126g
MW =
= 163.3 g mol
0.09260mol
(d) The calculated molecular weight is smaller than
true value, because:
measured g H2C2O4 is larger than true value,
(a) mol H2C2O4.2H2O =
1.2596 g
g
mol
=
calculated mol H2C2O4 is larger than true value,
calculated mol NaOH is larger than true value,
calculated M NaOH is larger than true value
calculated mol HX is larger than true value,
therefore,
g HX (true value )
MW =
mol HX (calculated , and too large )
1983 C
(a) Specify the properties of a buffer solution. Describe the components and the composition of
effective buffer solutions.
(b) An employer is interviewing four applicants for
a job as a laboratory technician and asks each
how to prepare a buffer solution with a pH close
to 9.
Archie A. says he would mix acetic acid and
sodium acetate solutions.
Beula B. says she would mix NH4Cl and HCl
solutions.
Carla C. says she would mix NH4Cl and NH3
solutions.
Dexter D. says he would mix NH3 and NaOH
solutions.
Which of these applicants has given an appropriate procedure? Explain your answer, referring to
your discussion in part (a). Explain what is
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teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
(b) Carla has the correct procedure, she has mixed a
weak base, NH3, with the salt of a weak base,
NH4Cl.
Archie has a buffer solution but a pH of around
5.
Beula doesn’t have a buffer solution, her solution consists of a strong acid and a salt of a weak
base.
Dexter does not have a buffer solution, since his
solution consists of a weak base plus a strong
base.
1984 A
Sodium benzoate, C6H5COONa, is the salt of a the
weak acid, benzoic acid, C6H5COOH. A 0.10 molar
solution of sodium benzoate has a pH of 8.60 at
room temperature.
(a) Calculate the [OH–] in the sodium benzoate
solution described above.
(b) Calculate the value for the equilibrium constant
for the reaction:
C6H5COO– + H2O √ C6H5COOH + OH–
(c) Calculate the value of Ka, the acid dissociation
constant for benzoic acid.
(d) A saturated solution of benzoic acid is prepared
by adding excess solid benzoic acid to pure
water at room temperature. Since this saturated
solution has a pH of 2.88, calculate the molar
solubility of benzoic acid at room temperature.
Answer:
(a) pH =8.6, pOH =5.4
[OH–] =10–pOH = 3.98×10–6M
(b) [C6H5COOH] = [OH–]
page 12
ions)
= 2.88×10–2M
1984 C
Discuss the roles of indicators in the titration of acids
and bases. Explain the basis of their operation and
the factors to be considered in selecting an
appropriate indicator for a particular titration.
Answer:
An indicator signals the end point of a titration
by changing color.
An indicator is a weak acid or weak base where
the acid form and basic form of the indicators are of
different colors.
An indicator changes color when the pH of the
solution equals the pKa of the indicator. In selecting
an indicator, the pH at which the indicator changes
color should be equal to (or bracket) the pH of the
solution at the equivalence point.
For example, when a strong acid is titrated with a
strong base, the pH at the equivalence point is 7, so
we would choose an indicator that changes color at a
pH = 7. {Many other examples possible.}
1986 A
In water, hydrazoic acid, HN3, is a weak acid that
has an equilibrium constant, Ka, equal to 2.8×10–5 at
25˚C. A 0.300 litre sample of a 0.050 molar solution
of the acid is prepared.
(a) Write the expression for the equilibrium
constant, Ka, for hydrazoic acid.
(b) Calculate the pH of this solution at 25˚C.
(c) To 0.150 litre of this solution, 0.80 gram of
sodium azide, NaN3, is added. The salt
−
−6
2
[C H COH][OH ]
(3.98 × 10 )
dissolved completely. Calculate the pH of the
Kb = 6 5
=
−
−6
resulting solution at 25˚C if the volume of the
[C6H 5CO2 ]
(0.1 − 3.98 × 10 )
solution remains unchanged.
= 1.58×10–10
(d) To the remaining 0.150 litre of the original soluKw
1.0 × 10 −14
−5
tion, 0.075 litre of 0.100 molar NaOH solution is
=
= 6.33 × 10
(c) Ka =
−10
Kb 1.58 × 10 )
added. Calculate the [OH–] for the resulting
solution at 25˚C.
(d) pH 2.88 = 1.32×10–3M [H+] = [C6H5COO–]
Answer:
[H+ ][C6 H5CO−2 ] (1.32 × 10 −3 )2
[C6H5CO2 H] =
=
(a) HN3 √ H+ + N3–
Ka
6.33 × 10−5
[H + ][N3− ]
= 2.75×10–2M [C6H5COOH]
Ka =
[HN3 ]
total dissolved = (2.75×10–2M + 1.32×10–3M as
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Acid - Base
(b) [H+] = [N3–] = X
X2
; X = 1.2 × 10 −3 M
0.050
+
pH = –log[H ] = 2.93
2.8 × 10−5 =
page 13
0.80g 1mol
×
= 0.082M
(c) [N−3 ] =
0.150L 65g
[H+ ](0.082)
2.8 × 10−5 =
; [H + ] = 1.7 × 10−5 M
0.050
pH = 4.77
(d) (0.075L)(0.100M) = 0.0075 mol NaOH
(0.150L)(0.050M) = 0.0075 mol HN3
OH– + HN3 → H2O + N3– ; neut. complete
K
N3– + H2O √ HN3 + OH– ; K b = w
Ka
1.0 × 10−14 [HN3 ][OH− ]
X2
=
=
 0.0075 
2.8 × 10−5
[N−3 ]
 0.225 
X = [OH–] = 3.5×10–6M
1986 D
H2SO3
HSO3–
HClO4
HClO3
H3BO3
Oxyacids, such as those above, contain an atom
bonded to one or more oxygen atoms; one or more of
these oxygen atoms may also be bonded to hydrogen.
(a) Discuss the factors that are often used to predict
correctly the strengths of the oxyacids listed
above.
(b) Arrange the examples above in the order of increasing acid strength.
Answer:
(a) 1) As effective nuclear charge on central atom
increases, acid strength increases. OR
As number of lone oxygen atoms (oxygen atoms
not bonded to hydrogen) increases, acid strength
increases. OR
As electronegativity of central atom increases,
acid strength increases.
2) Loss of H+ by a neutral acid molecule
reduces acid strength. OR
Ka of H2SO3 > Ka of HSO3–
(b) H3BO3 < HSO3– < H2SO3 < HClO3 < HClO4
H3BO3 or HSO3– weakest (must be together)
1987 A
NH3 + H2O √ NH4+ + OH– Ammonia is a weak base
that dissociates in water as shown above. At 25˚C,
the base dissociation constant, Kb, for NH3 is
1.8×10–5.
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teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
(a) Determine the hydroxide ion concentration and
the percentage dissociation of a 0.150 molar solution of ammonia at 25˚C.
(b) Determine the pH of a solution prepared by
adding 0.0500 mole of solid ammonium chloride
to 100. millilitres of a 0.150 molar solution of
ammonia.
(c) If 0.0800 mole of solid magnesium chloride,
MgCl2, is dissolved in the solution prepared in
part (b) and the resulting solution is well–stirred,
will a precipitate of Mg(OH)2 form? Show
calculations to support your answer. (Assume
the volume of the solution is unchanged. The
solubility product constant for Mg(OH)2 is
1.5×10–11.
Answer:
(a) [NH4+] = [OH–] = X; [NH3] = (0.150 – X)
page 14
in water and titrated with the NaOH solution. To
reach the equivalence point, 26.90 millilitres of
base was required. Calculate the molarity of the
NaOH solution. (Molecular weight: KHC8H4O4
= 204.2)
X2
[NH+4 ][OH− ]
–5
Kb =
; 1.8×10 =
0.150 – X
[NH3]
X= [OH–] = 1.6×10–3 M
1.6 ×103
% diss. =
× 100% = 1.1%
0.150
(b) [NH4+] = 0.0500 mol/0.100L = 0.500M
[NH3] = 0.150M
(0.500)(X)
1.8 × 10−5 =
; X = [OH− ] = 5.4 × 10−6 M
0.150
pOH = 5.27; pH = (14 – 5.27) = 8.73
(c) Mg(OH)2 √ Mg2+ + 2 OH–
[Mg2+] = (0.0800mol/0.100L) = 0.800M
[OH–] = 5.4×10–6M
Q = [Mg2+][OH–]2 = (0.800)(5.4×10–6)2
= 2.3×10–11
Q > Ksp so Mg(OH)2 precipitates
1987 B
The percentage by weight of nitric acid, HNO3, in a
sample of concentrated nitric acid is to be
determined.
(a) Initially a NaOH solution was standardized by
titration with a sample of potassium hydrogen
phthalate, KHC8H4O4, a monoprotic acid often
used as a primary standard. A sample of pure
KHC8H4O4 weighing 1.518 grams was dissolved
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teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
(b) A 10.00 millilitre sample of the concentrated nitric acid was diluted with water to a total volume
of 500.00 millilitres. Then 25.00 millilitres of the
diluted acid solution was titrated with the standardized NaOH solution prepared in part (a). The
equivalence point was reached after 28.35 millilitres of the base had been added. Calculate the
molarity of the concentrated nitric acid.
(c) The density of the concentrated nitric acid used
in this experiment was determined to be 1.42
grams per millilitre. Determine the percentage
by weight of HNO3 in the original sample of
concentrated nitric acid.
Answer:
1mol
= 7.434 × 10 −3 mol acid
(a) 1.518g ×
204.2g
= mol NaOH required to neut.
page 15
1988 D
12
X
10
X
8
X
X
pH 6
4X
X
X
X
2
0
0
5
10 15 20 25 30
millilitres of NaOH
A 30.00 millilitre sample of a weak monoprotic acid
was titrated with a standardized solution of NaOH. A
pH meter was used to measure the pH after each in−3
crement of NaOH was added, and the curve above
7.434 × 10 mol
= 0.2764M NaOH
was constructed.
0.02690L
28.35mLNaOH 0.2764mol 1molHNO3
(a) Explain how this curve could be used to deter×
×
=
(b)
mine the molarity of the acid.
25.00mLHNO3
1L
1molNaOH
(b) Explain how this curve could be used to deter= 0.3134M HNO3
mine the dissociation constant Ka of the weak
MfVf=MiVi;(0.3134M)(500mL)
=
monoprotic acid.
(M)(10.00mL)
(c) If you were to repeat the titration using a
M = 15.67M HNO3
indicator in the acid to signal the endpoint,
g L HNO3
which of the following indicators should you
× 100%
(c) % HNO3 in conc. sol’n =
g L sol’n
select? Give the reason for your choice.
grams HNO3 in 1 L conc. sol’n =
Methyl red
Ka = 1×10–5
15.67molHNO3 63.02g
Cresol red
Ka = 1×10–8
g
×
= 987.5 L
1L
1 mol
Alizarin yellow
Ka = 1×10–11
grams sol’n in 1 L conc. sol’n
(d) Sketch the titration curve that would result if the
1.42g sol’n 1000 mL
g
weak monoprotic acid were replaced by a strong
×
=1420 L
1 mL
1L
monoprotic acid, such as HCl of the same
mol
molarity. Identify differences between this
(0.01523 L)(0.2211 L )
[HA] =
= 0.04810M
titration curve and the curve shown above.
0.07000 L
Answer:
[OH− ][HA] Kw 1. ×10−14
−10
K=
=
=
=1.3 ×10
(a) The sharp vertical rise in pH on the pH–volume
Ka 7.7 ×10−5
[A− ]
curve appears at the equivalence point (about 23
−3
mL). Because the acid is monoprotic, the
7.789 ×10 mol
[A − ] =
= 0.0914M
number of moles of acid equals the number of
0.08523 L
moles of NaOH. That number is the product of
the exact volume and the molarity of the NaOH.
The molarity of the acid is the number of moles
of the acid divided by 0.30L, the volume of the
acid.
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teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
page 16
(b) At the half–equivalence point (where the
ascorbic acid.
volume of the base added is exactly half its (c) Calculate the equilibrium constant for the
volume at the equivalence point), the
reaction of the ascorbate ion, A–, with water.
concentration [HX] of the weak acid equals the (d) Calculate the pH of the solution at the
concentration [X–] of its anion. Thus, in the
equivalence point of the titration.
equilibrium expression [H+][X–]/[HX] = Ka,
[H+] = Ka. Therefore, pH at the half–
equivalence point equals pKa.
(c) Cresol red is the best indicator because its pKa
(about 8) appears midway in the steep equivalence region. This insures that at the equivalence
point the maximum color change for the
minimal change in the volume of NaOH added
is observed.
(d)
12
10
8
pH 6
Longer equivalence
region for strong acid
X
X
X
X
X
X
X
Same
volume
of
NaOH
4X
added
for equivalence point
2
Acid portion at lower
0
pH for strong acid
0
5
10 15 20 25 30
millilitres of NaOH
1989 A
In an experiment to determine the molecular weight
and the ionization constant for ascorbic acid (vitamin
C), a student dissolved 1.3717 grams of the acid in
water to make 50.00 millilitres of solution. The
entire solution was titrated with a 0.2211 molar
NaOH solution. The pH was monitored throughout
the titration. The equivalence point was reached
when 35.23 millilitres of the base has been added.
Under the conditions of this experiment, ascorbic
acid acts as a monoprotic acid that can be
represented as HA.
(a) From the information above, calculate the
molecular weight of ascorbic acid.
(b) When 20.00 millilitres of NaOH had been added
during the titration, the pH of the solution was
4.23. Calculate the acid ionization constant for
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teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
Answer:
(a) (0.2211M)(0.03523L) = 7.789×10–3 mol
1.3717g/7.789×10–3 mol = 176.1g/mol
(b) at pH 4.23, [H+] = 8.0×10–8M
page 17
to its greater charge density. This draws
electrons in the H–O bond towards it and
weakens the bond. H+ can be removed, making
an acidic solution.
(c) Water is a more basic solvent (greater attraction
(0.02000L)(0.2211mol ⋅ L−1 )
for H+) and removes H+ from HCl and HI
[A − ] =
= 0.06317M
0.07000L
equally.
(0.01523L)(0.2211mol ⋅ L−1 )
Acetic acid has little attraction for H+, but the
[HA] =
= 0.04810M
H+ separates from the larger I– more easily than
0.07000L
+
−
−5
from the smaller Cl–.
[H ][A ] (5.9 × 10 )(0.06317)
−5
K=
=
= 7.7 × 10
(d) The bond between H and Cl is weaker than the
[HA]
(0.04810)
bond between H and F. Therefore, HCl is a
(c) A– + H2O √ HA + OH–
stronger acid.
[OH− ][HA] Kw 1.×10 −14
K=
=
=
= 1.3 × 10 −10
−
−5
1991 A
K
[A ]
7.7 × 10
a
The acid ionization constant, Ka, for propanoic acid,
(d) at equiv. pt.
C2H5COOH, is 1.3×10–5.
7.789 × 10 −3 mol
−
[A ] =
= 0.0914M
(a) Calculate the hydrogen ion concentration, [H+],
0.08523L
in a 0.20–molar solution of propanoic acid.
[OH–]2 = (1.3×10–10)(9.14×10–2) = 1.2×10–11
(b) Calculate the percentage of propanoic acid
[OH–] = 3.4×10–6M
molecules that are ionized in the solution in (a).
pOH =–log(3.4×10–6) =5.47; pH = (14–5.47)=
(c) What is the ratio of the concentration of
8.53
propanoate ion, C2H5COO–, to that of propanoic
acid in a buffer solution with a pH of 5.20?
1990 D
(d) In a 100.–milliliter sample of a different buffer
Give a brief explanation for each of the following.
solution, the propanoic acid concentration is
(a) For the diprotic acid H2S, the first dissociation
0.35–molar and the sodium propanoate concenconstant is larger than the second dissociation
tration is 0.50–molar. To this buffer solution,
constant by about 105 (K1 ≈ 105 K2).
0.0040 mole of solid NaOH is added. Calculate
(b) In water, NaOH is a base but HOCl is an acid.
the pH of the resulting solution.
(c) HCl and HI are equally strong acids in water Answer:
but, in pure acetic acid, HI is a stronger acid
[H+ ][C2 H5 COO]
(a)
= Ka
than HCl.
[C2 H5 COOH]
(d) When each is dissolved in water, HCl is a much
[H+] = [C2H5COO–] = X
stronger acid than HF.
[C2H5COOH] = 0.20 M – X
Answer:
(a) After the first H+ is lost from H2S, the
assume X is small, ∴, 0.20 – X ≈ 0.20
–
remaining species, HS , has a negative charge.
X2
= 1.3×10−5 ;X =1.6×10−3 M = [H+ ]
This increases the attraction of the S atom for
0.20
the bonding electrons in HS–. Therefore, the
(b) from (a), X = amount of acid that ionized, ∴,
bond is stronger, H+ is harder to remove, and K2
1.6 ×103
is lower.
× 100% = 0.81% ionized
0.20
+
(b) Polar H2O can separate ionic NaOH into Na (aq)
(c) @ pH 5.20, [H+] = antilog (–5.20) = 6.31×10–6
and OH–(aq), giving a basic solution. In HOCl,
M
chlorine has a high attraction for electrons due
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teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
(6.3×10−6 )[C2H5COO− ]
[C2H5COOH]
[C2 H5 COO]
2.1
=
[C2 H5 COOH] 1
OR
= Ka = 1.3×10−5
page 18
(c) Henderson–Hasselbalch
[base]
pH = pK a + log
[acid]
-
[C 3H 5O2]
5.20 = 4.89 + log
[HC 3H 5O2]
-
[C 3H 5O2]
log
= 0.31 = 2.1
[HC 3H 5O2]
(d) [C2H5COO–] = 0.50 M; [C2H5COOH] = 0.35 M
[OH–] = 0.0040 mol/0.100 L = 0.040 M
this neutralizes 0.04 M of the acid, giving
[C2H5COOH] = 0.31 M and the propanoate ion
increases by a similar amount to 0.54 M.
+
[H ](0.54)
-5
+
-6
= 1.3 x 10 , [H ] = 7.5 x 10 M
0.31
pH = – log [H+] = 5.13
OR
(d) using [ ]’s or moles of propanoic acid and
propanoate ion...
0.54
pH = pK a + log
0.31
= 4.89 + 0.24 = 5.13
1992 D
The equations and constants for the dissociation of
three different acids are given below.
HCO3– √ H+ + CO32–
Ka = 4.2 × 10–7
H2PO4– √ H+ + HPO42–
Ka = 6.2 × 10–8
HSO4– √ H+ + SO42–
Ka = 1.3 × 10–2
(a) From the systems above, identify the conjugate
pair that is best for preparing a buffer with a pH
of 7.2. Explain your choice.
(b) Explain briefly how you would prepare the
buffer solution described in (a) with the
conjugate pair you have chosen.
(c) If the concentrations of both the acid and the
conjugate base you have chosen were doubled,
how would the pH be affected? Explain how the
capacity of the buffer is affected by this change
in concentrations of acid and base.
(d) Explain briefly how you could prepare the
buffer solution in (a) if you had available the
solid salt of the only one member of the
conjugate pair and solution of a strong acid and
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teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
a strong base.
page 19
Answer:
(a) Best conjugate pair: H2PO4–, HPO42–. When
7.2 = pH = pKa for this pair when [H2PO4–] =
[HPO42–].
(b) Dissolve equal moles (or amounts) of H2PO4–,
and HPO42– (or appropiate compounds) in
water.
(c) pH not changed. Capacity of buffer would increase because there are more moles of
conjugate acid and conjugate base to react with
added base or acid.
(d) Add strong base to salt of conjugate acid OR
add strong acid to salt of conjugate base.
Add 1 mole conjugate acid to 1/2 mole strong
base OR 1 mole conjugate base to 1/2 mole
strong acid.
OR
Use pH meter to monitor addition of strong base
to conjugate acid OR strong acid to conjugate
base.
1993 A
CH3NH2 + H2O √ CH3NH3+ + OH–
Methylamine, CH3NH2, is a weak base that reacts
according to the equation above. The value of the
ionization constant, Kb, is 5.25×10–4. Methylamine
forms salts such as methylammonium nitrate,
(CH3NH3+)(NO3–).
(a) Calculate the hydroxide ion concentration, [OH–
], of a 0.225–molar aqueous solution of methylamine.
(b) Calculate the pH of a solution made by adding
0.0100 mole of solid methylammonium nitrate
to 120.0 milliliters of a 0.225–molar solution of
methylamine. Assume no volume change
occurs.
(c) How many moles of either NaOH or HCl (state
clearly which you choose) should be added to
the solution in (b) to produce a solution that has
a pH of 11.00? Assume that no volume change
occurs.
(d) A volume of 100. milliliters of distilled water is
added to the solution in (c). How is the pH of
the solution affected? Explain.
Answer:
Copyright © 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroom
teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
(a) Kb =
[ ]i
∆[ ]
[ ]eq
0.0227 M × 0.120 L =
[CH3NH+3 ][OH− ]
[CH3NH2 ]
CH3NH2 + H2O √ CH3NH3+ + OH–
0.225
0
0
–X
+X
+X
0.225–X
X
X
[X][X]
X2
Kb = 5.25×10 =
≅
[0.225 − X] 0.225
X = [OH–] = 1.09×10–2 M
solved using quadratic: X = [OH–] = 1.06×10–2
M
(b) [CH3NH3+] = 0.0100 mol / 0.1200 L = 0.0833
M
or CH3NH2 = 0.120 L × 0.225 mol/L = 0.0270
mol
[0.0833 + X][X] 0.0833X
Kb = 5.25×10−4 =
≅
[0.225 − X]
0.225
–
–3
X = [OH ] = 1.42×10 M; pOH = 2.85; pH =
11.15
OR
[base]
pH = pKa + log
[acid]
−4
Ka =
1×10−14
− 11
−4 = 1.91×10
5.25×10
pH = 10.72 + log
; pKa =10.72
(0.225)
=11.15
(0.0833)
OR
[acid]
; pKb = 3.28
[base]
(0.0833)
pOH = 3.28 + log
= 2.85; pH =11.15
(0.225)
(c) HCl must be added.
[0.0833 + X][0.0010]
Kb = 5.25×10−4 =
[0.225 − X]
X = 0.0228 M
0.0228 mol/L × 0.120 L = 2.74×10–3 mol HCl
OR
[base]
[base]
11.00 = 10.72 + log
; log
= 0.28
[acid]
[acid]
[base]
(0.225 − X)
=1.905 =
; X = 0.0227M
[acid]
(0.0833 + x)
pOH = pKb + log
page 20
(d) The
2.73×10–3
mol HCl
[CH3 NH+3 ]
ratio does not change in this
[CH3 NH2 ]
buffer solution with dilution, therefore, no effect
on pH.
1993 D (Required)
The following observations are made about reaction
of sulfuric acid, H2SO4. Discuss the chemical
processes involved in each case. Use principles from
acid–base theory, oxidation–reduction, and bonding
and/or intermolecular forces to support your answers.
(a) When zinc metal is added to a solution of dilute
H2SO4, bubbles of gas are formed and the zinc
disappears.
(b) As concentrated H2SO4 is added to water, the
temperature of the resulting mixture rises.
(c) When a solution of Ba(OH)2 is added to a dilute
H2SO4 solution, the electrical conductivity decreases and a white precipitate forms.
(d) When 10 milliliters of 0.10–molar H2SO4 is
added to 40 milliliters of 0.10–molar NaOH, the
pH changes only by about 0.5 unit. After 10
more milliliters of 0.10–molar H2SO4 is added,
the pH changes about 6 units.
Answer:
(a) Zn is oxidized to Zn2+ by H+ which in turn is reduced by Zn to H2. Identify H2(g) or Zn dissolving as Zn2+.
Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)
Explicit: Redox or e– transfer or correctly
identify oxidizing agent or reducing agent.
(b) H2SO4 dissociates, forms ions or dydration
“event”. Bonds form, therefore, energy given off
(connection).
(c) BaSO4 (ppt) forms or H+ + OH– form water.
Newly formed water and ppt remove ions lowering conductivity.
Ba2+(aq) + OH–(aq) + H+(aq) + SO42–(aq) →
BaSO4(s)+ H2O(l)
(d) First 10 mL produces solution of SO42– and
OH– or excess OH– or partial neutralization (pH
13.0 → 12.6). [presence of HSO4– in solution voids this
point]
Copyright © 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroom
teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
Second 10 mL produces equivalence where pH
decreases (changes) rapidly (pH 12.6 → 7.0). [pH
3–
“rises” or wrong graph, if used, voids this point]
1994 D
A chemical reaction occurs when 100. milliliters of
0.200–molar HCl is added dropwise to 100.
milliliters of 0.100–molar Na3P04 solution.
(a) Write the two net ionic equations for the formation of the major products.
(b) Identify the species that acts as both a Brönsted
acid and as a Brönsted base in the equation in
(a), Draw the Lewis electron–dot diagram for
this species.
(c) Sketch a graph using the axes provided, showing
the shape of the titration curve that results when
100. milliliters of the HCl solution is added
slowly from a buret to the Na3PO4 solution. Account for the shape of the curve.
pH
0
mL HCl
(d) Write the equation for the reaction that occurs if
a few additional milliliters of the HCl solution
are added to the solution resulting from the
titration in (c).
Answer:
(a) PO43– + H+ → HPO42–; HPO42– + H+ → H2PO4–
. .–
:
. .:. .
.. O
O
. . ::.P. : .O. : H
:O
. .–:
(b) HPO42–,
page 21
+
2–
PO 4+ H → HPO4
pH
–
HPO42–+ H +→ H2PO4
0
mL HCl
(c)
(d) H+ + H2PO4– → H3PO4
1996 A
HOCl √ OCl– + H+
Hypochlorous acid, HOCl, is a weak acid commonly
used as a bleaching agent. The acid-dissociation constant, Ka, for the reaction represented above is
3.2×10–8.
(a) Calculate the [H+] of a 0.14-molar solution of
HOCl.
(b) Write the correctly balanced net ionic equation
for the reaction that occurs when NaOCl is
dissolved in water and calculate the numerical
value of the equilibrium constant for the
reaction.
(c) Calculate the pH of a solution made by combining 40.0 milliliters of 0.14-molar HOCl and 10.0
milliliters of 0.56-molar NaOH.
(d) How many millimoles of solid NaOH must be
added to 50.0 milliliters of 0.20-molar HOCl to
obtain a buffer solution that has a pH of 7.49?
Assume that the addition of the solid NaOH results in a negligible change in volume.
(e) Household bleach is made by dissolving
chlorine gas in water, as represented below.
Cl2(g) + H2O → H+ + Cl– + HOCl(aq)
Calculate the pH of such a solution if the concentration of HOCl in the solution is 0.065 molar.
Answer:
[OCl – ][H+ ]
(a) Ka =
= 3.2×10–8
[HOCl]
X = amount of acid that ionizes = [OCl–] = [H+]
(0.14 – X) = [HOCl] that remains unionized
X2
=
; X = 6.7×10–5 M = [H+]
0.14 – X
(b) NaOCl(s) + H2O → Na+(aq) + HOCl(aq) + OH–
3.2×10–8
Copyright © 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroom
teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
(aq)
14
K w 1 ×10
=
= 3.1×10–7
K a 3.2 ×108
(c) [ ]o after dilution but prior to reaction:
40 mL
[HOCl] = 0.14 M ×
= 0.11 M
50mL
10 mL
[OH–] = 0.56 M ×
= 0.11 M
50mL
Equivalence point reached. [OH–] ≈ [HOCl]
Kb =
page 22
(e) 1 mol H+ for every 1 mole of HOCl produced
[H+] ≈ [HOCl] = 0.065 M
pH = – log (0.065) = 1.2
1997 A
The overall dissociation of oxalic acid, H2C2O4, is
represented below. The overall dissociation constant
is also indicated.
H2C2O4 √ 2 H+ + C2O42–
K = 3.78×10–6
(a) What volume of 0.400-molar NaOH is required
[OH– ]2
–7
to neutralize completely a 5.00×10–3-mole
Kb =
= 3.1×10
0.11
sample of pure oxalic acid?
[OH–] = 1.8×10–4 ; pOH = 3.7
(b) Give the equations representing the first and secpH = 14 – 3.7 = 10.3
ond dissociations of oxalic acid. Calculate the
(d) at pH 7.49, the [H+] = 10–7.49 = 3.24×10–8 M
value of the first dissociation constant, K1, for
oxalic acid if the value of the second
when the solution is half-neutralized, pH = pKa
–
dissociation constant, K2, is 6.40×10–5.
[OCl ]
and
=1
(c) To a 0.015-molar solution of oxalic acid, a
[HOCl]
strong acid is added until the pH is 0.5.
0.20 mol HOCl
× 50.0 mL = 10.0 mmol HOCl
Calculate the [C2O42–] in the resulting solution.
1L
(Assume the change in volume is negligible.)
half this amount, or 5.0 mmol of NaOH added.
(d) Calculate the value of the equilibrium constant,
Kb, for the reaction that occurs when solid
Na2C2O4 is dissolved in water.
Answer:
2 mol H+
(a)
mol oxalic acid ×
×
1 mol oxalic acid
1 mol OH 1000. mL NaOH
×
=
0.400 mol NaOH
1 mol H+
= 25.0 mL NaOH
(b) H2C2O4 √ H+ + HC2O4–
5.00×10–3
HC2O4– √ H+ + C2O42–
K = K1 × K2
3.78 ×106
K
=
= 5.91×10–2
K2
6.40 ×105
(c) X = amt. ionized
[H2C2O4] = 0.015 – X
K1 =
[H+] = 10–pH = 10–0.5 = 0.316 M
[C2O42–] = X
Ka =
[H+ ]2 [C2 O4 2 ]
[H2 C2 O4 ]
= 3.78×10–6
Copyright © 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroom
teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
2
3.78×10–6 =
[0.316] [X]
; X = 5.67×10–7 M
[0.015 – X]
1 ×101 4
Kw
(d) Kb =
=
= 1.56×10–10
K2
6.40 ×106
page 23
(d) Describe how the value of the acid-dissociation
constant, Ka, for the weak acid HX could be determined from the titration curve in part (c).
1998 D (Required)
[repeated in lab procedures section]
An approximately 0.1-molar solution of NaOH is to
be standardized by titration. Assume that the following materials are available.
• Clean, dry 50 mL buret
• 250 mL Erlenmeyer flask
• Wash bottle filled with distilled water
• Analytical balance
• Phenolphthalein indicator solution
• Potassium hydrogen phthalate, KHP, a pure
solid monoprotic acid (to be used as the primary
standard)
(a) Briefly describe the steps you would take, using
the materials listed above, to standardize the
NaOH solution.
(b) Describe (i.e., set up) the calculations necessary
to determine the concentration of the NaOH
solution.
(c) After the NaOH solution has been standardized,
it is used to titrate a weak monoprotic acid, HX.
The equivalence point is reached when 25.0 mL
of NaOH solution has been added. In the space
provided at the right, sketch the titration curve,
showing the pH changes that occur as the
volume of NaOH solution added increases from
0 to 35.0 mL. Clearly label the equivalence
point on the curve.
Copyright © 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroom
teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
page 24
10pKa
(e) The graph below shows the results obtained by
pKa. Ka =
titrating a different weak acid, H2Y, with the (e) Y2– (could it be OH– ?)
standardized NaOH solution. Identify the negative ion that is present in the highest
1998 D
concentration at the point in the titration
Answer each of the following using appropriate
represented by the letter A on the curve.
chemical principles.
(b) When NH3 gas is bubbled into an aqueous solution of CuCl2, a precipitate forms initially. On
further bubbling, the precipitate disappears. Explain these two observations.
In each case, justify your choice.
Answer
(b) A small amount of NH3 in solution causes an increase in the [OH–].
Answer
NH3 + H2O √ NH4+ + OH–
(a) • exactly mass a sample of KHP in the Erlenmeyer
flask and add distilled water to dissolve the solid.
This, in turn, causes the Ksp of copper(II)
• add a few drops of phenolphthalein to the flask.
hydroxide to be exceeded and the solution forms
a precipitate of Cu(OH)2.
• rinse the buret with the NaOH solution and fill.
• record starting volume of base in buret.
With the addition of more NH3, you form the
• with mixing, titrate the KHP with the NaOH
soluble tetraamminecopper(II) complex ion,
solution until it just turns slightly pink.
[Cu(NH3)4]2+, which will cause the precipitate
• record end volume of buret.
to dissolve.
• repeat to check your results.
mass of KHP
1999 A
(b)
= moles of KHP
molar mass KHP
NH3(aq) + H2O(l) √ NH4+(aq) + OH–(aq)
since KHP is monoprotic, this is the number of In aqueous solution, ammonia reacts as represented
moles of NaOH
above. In 0.0180 M NH3(aq) at 25˚C, the hydroxide
moles of NaOH
ion concentration, [OH–] is 5.60×10–4 M. In answer= molarity of NaOH
L of titrant
ing the following, assume that temperature is
constant at 25˚C and that volumes are additive.
(a) Write the equilibrium-constant expression for
the reaction represented above.
equivalence point
(b) Determine the pH of 0.0180 M NH3(aq).
(c) Determine the value of the base ionization
constant, Kb, of NH3(aq).
(d) Determine the percent ionization of NH3 in
0.0180 M NH3(aq).
(e) In an experiment, a 20.0 mL sample of 0.0180
M NH3(aq) was placed in a flask and titrated to
the equivalence point and beyond using 0.0120
M HCl(aq).
(i) Determine the volume of 0.0120 M HCl(aq)
(c)
that was added to reach the equivalence
(d) from the titration curve, at the 12.5 mL volume
point.
point, the acid is half-neutralized and the pH =
Copyright © 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroom
teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
Acid - Base
(ii) Determine the pH of the solution in the
flask after a total of 15.0 mL of 0.0120 M
HCl(aq) was added.
(iii) Determine the pH of the solution in the
flask after a total of 40.0 mL of 0.0120 M
HCl(aq) was added.
Answer
[NH4+][OH–]
(a) Kb =
[NH3]
(b) pOH = -log(5.60×10–4) = 3.252
pH = 14 – pOH = 10.748
(5.60∞10–4)(5.60∞10–4)
(c) Kb =
= 1.80×10–5
(0.0180 – 5.60∞10–4)
5.60∞10–4
0.0180 × 100% = 3.11%
(e) (i) NAVA = NBVB
(0.0120 N)(VA) = (0.0180 N)(20.0 mL)
VA = 30.0 mL
(ii) at 15.0 mL it is half-titrated and [NH4+] =
[NH3], then the Kb = [OH–] = 1.80×10–5.
(d)
pOH = -log(1.80×10–5) = 4.745
pH = 14 – 4.75 = 9.255
(iii) at 40.0 mL, there is an excess of 10.0 mL
of HCl past equivalence point,
(10.0 mL)(0.0120 M) = 0.120 mmol H+ ions
0.120 mmol
+
60.0 mL = 0.00200M =[H ]
pH = -log(0.00200) = 2.70
Copyright © 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroom
teachers are permitted to reproduce the questions. Portions opyright © 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.
page 25